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CHAPTER 6 Mechanical Properties Of Metals - I 6-1.

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1 CHAPTER 6 Mechanical Properties Of Metals - I 6-1

2 Processing of Metals - Casting Most metals are first melted in a furnace. Alloying is done if required. Large ingots are then cast. Sheets and plates are then produced from ingots by rolling Wrought alloy products. Channels and other shapes are produced by extrusion. Some small parts can be cast as final product. Example :- Automobile Piston. 6-2

3 Casting (Cont..) Casting Process Casting mold Cast parts Figure 5.3 a Figure 5.3 b Figure

4 Hot Rolling of Steel Hot rolling Greater reduction of thickness in a single pass. Rolling carried out above recrystallization temperature. Ingots preheated to about C. Ingots reheated between passes if required. Usually, series of 4 high rolling mills are used. 6-4

5 Cold Rolling of Metal Sheet Cold rolling is rolling performed below recrystallization temperature. This results in strain hardening. Hot rolled slabs have to be annealed before cold rolling. Series of 4 high rolling mills are usually used. Less reduction of thickness. Needs high power. Figure

6 Cold Rolling (Cont..) % Cold work = Initial metal thickness – Final metal thickness Initial metal thickness x 100 Figure

7 Extrusion Metal under high pressure is forced through opening in a die. Common Products are cylindrical bar, hollow tubes from copper, aluminum etc. Normally done at high temperature. Indirect extrusion needs less power however there is a limit on load applied Container Metal Die Direct Extrusion indirect Extrusion 6-7 Figure 5.9

8 Forging Metal, usually hot, is hammered or pressed into desired shape. Types:- Open die: Dies are flat and simple in shape * Example products: Steel shafts Closed die: Dies have upper and lower impresion * Example products: Automobile engine connection rod. Forging improves structural properties, removes porosity and increases homogeneity. Direct Forging Indirect Forging Dies Metal 6-8

9 Drawing Wire drawing :- Starting rod or wire is drawn through several drawing dies to reduce diameter. Deep drawing:- Used to shape cup like articles from flats and sheets of metals % cold work = Change in cross-sectional area Original area X 100 Wire or rod Carbide nib Figure 5.13 Figure

10 Stress and Strain in Metals Metals undergo deformation under uniaxial tensile force. Elastic deformation: Metal returns to its original dimension after tensile force is removed. Plastic deformation: The metal is deformed to such an extent that it cannot return to its original dimensions Elastic deformation Plastic deformation 6-10

11 Engineering Stress and Strain Engineering stress σ = A0A0 Δl A Figure 5.15 F (Average uniaxial tensile force) A 0 (Original cross-sectional area) Units of Stress are PSI or N/M 2 (Pascals) 1 PSI = 6.89 x 10 3 Pa Engineering strain = ε = Change in length Original length Units of strain are in/in or m/m. 6-11

12 Poisons Ratio w0w0 w. Poisons ratio = Usually poisons ratio ranges from 0.25 to 0.4. Example: Stainless steel 0.28 Copper

13 Shear Stress and Shear Strain τ = Shear stress = V (Shear force) A (Area of shear force application) Shear strain γ = Amount of shear displacement Distance h over which shear acts. Elastic Modulus G = τ / γ Figure

14 Tensile test Strength of materials can be tested by pulling the metal to failure. Specimen Extensometer Force data is obtained from Load cell Strain data is obtained from Extensometer. Load Cell Figure

15 Tensile Test (Cont) Commonly used Test specimen Typical Stress-strain curve Figure 5.21 Figure

16 Mechanical Properties Modulus of elasticity (E) : Stress and strain are linearly related in elastic region. (Hooks law) Higher the bonding strength, higher is the modulus of elasticity. Examples: Modulus of Elasticity of steel is 207 Gpa. Modulus of elasticity of Aluminum is 76 Gpa E = σ (Stress) ε (Strain) Stress Strain Linear portion of the stress strain curve Δε Δσ E = Δσ Δε 6-17

17 Yield Strength Yield strength is strength at which metal or alloy show significant amount of plastic deformation. 0.2% offset yield strength is that strength at which 0.2% plastic deformation takes place. Construction line, starting at 0.2% strain and parallel to elastic region is drawn to fiend 0.2% offset yield strength. Figure

18 Ultimate tensile strength Ultimate tensile strength (UTS) is the maximum strength reached by the engineering stress strain curve. Necking starts after UTS is reached. More ductile the metal is, more is the necking before failure. Stress increases till failure. Drop in stress strain curve is due to stress calculation based on original area. Al 2024-Annealed Al 2024-Tempered S T R E S Mpa Strain Necking Point Stress strain curves of Al 2024 With two different heat treatments. Ductile annealed sample necks more Figure

19 Percent Elongation Percent elongation is a measure of ductility of a material. It is the elongation of the metal before fracture expressed as percentage of original length. % Elongation = Measured using a caliper fitting the fractured metal together. Example:- Percent elongation of pure aluminum is 35% For 7076-T6 aluminum alloy it is 11% Final length – initial Length Initial Length 6-20

20 Percent Reduction in Area Percent reduction area is also a measure of ductility. The diameter of fractured end of specimen is meas- ured using caliper. Percent reduction in area in metals decreases in case of presence of porosity. % Reduction Area = Initial area – Final area Final area Figure 5.14 Stress-strain curves of different metals 6-21

21 True Stress – True Strain True stress and true strain are based upon instantaneous cross-sectional area and length. True Stress = σ t = True Strain = ε t = True stress is always greater than engineering stress. F A i (instantaneous area) 6-22

22 Hardness and Hardness Testing Hardness is a measure of the resistance of a metal to permanent (plastic) deformation. General procedure: Press the indenter that is harder than the metal Into metal surface. Withdraw the indenter Measure hardness by measuring depth or width of indentation. Figure 5.27 Rockwell hardness tester 6-27

23 Hardness Tests Table

24 Plastic Deformation in Single Crystals Plastic deformation of single crystal results in step markings on surface slip bands. Atoms on specific crystallographic planes (slip planes) slip to cause slip bands. Figure 5.28 Slip bands 6-29

25 Slip Bands and Slip Planes Slip bands in ductile metals are uniform (occurs in many slip planes). Slip occurs in many slip planes within slip bands. Slip planes are about 200A thick and are offset by about 2000A Figure

26 Slip Mechanism During shear, atoms do not slide over each other. The slip occurs due to movement of dislocations. Dislocation cell structure in lightly deformed Aluminum Figure 5.33 Figure 5.32 Wall of high dislocation density 6-31

27 Slip in Crystals Slip occurs in densely or close packed planes. Lower shear stress is required for slip to occur in densely packed planes. If slip is restricted in close planes, then less dense planes become operative. Less energy is required to move atoms along denser planes. Close packed plane Non-close-packed plane Figure

28 Slip Systems Slip systems are combination of slip planes and slip direction. Each crystal has a number of characteristic slip systems. In FCC crystal, slip takes place in {111} octahedral planes and directions. 4 (111) type planes and 3 [110] type directions. 4 x 3 = 12 slip systems. Table

29 Slip Systems in BCC Crystal BCC crystals are not close packed. The slip predominantly occurs in {110} planes that has highest atomic density. Table

30 Slip Systems in HCP Crystal If HCP crystals have high c/a ratio, slip occurs along basal planes {0001}. For crystals with low c/a ratio, slip also occurs in {1010} and {1011} planes. Table

31 Critical Resolved Shear Stress Critical resolved shear stress is the stress required to cause slip in pure metal single crystal. Depends upon Crystal Structure Atomic bonding characteristics Temperature Orientation of slip planes relative to shear stress Slip begins when shear stress in slip plane in slip direction reaches critical resolved shear stress. This is equivalent to yield stress. Example :- Zn HCP % pure 0.18MPa Ti HCP 99.99% pure 13.7 MPa Ti HCP 99.9% pure 90.1 MPa 6-36

32 Schmids Law The relationship between uniaxial stress action on a single cylinder of pure metal single crystal and resulting resolved shear stress produced on a slip system is give by τr=τr= τr=τr= Shear Force Shear Area Slip direction Normal to Slip plane A 1 =Area of Slip plane 6-37

33 Twinning In twinning, a part of atomic lattice is deformed and forms mirror image of lattice next to it. Distance moved by atoms is proportional to their distance from twinning plane. Deformation from twinning is small. Twinning reorients the slip system. Twining is important for deformation in HCP crystals due to lesser slip planes. 6-38

34 Effects of Grain Boundaries on Strength Grain boundaries stop dislocation movement and hence strengthen the metals. Fine grain size is desirable, and hence metals are produced with finer grains. Figure 5.40 Stress-strain curve of single and polycrystalline copper Slip bands in polycrystalline aluminum grains Dislocations piled up against grain boundaries in stainless steel 6-39

35 Hall Petch Equation The Finer the grains, the better the mechanical properties (at room temperature). More isotropic properties Less resistant to corrosion and creep Hall-Petch equation - Empirical y = o + k / (d) 1/2 y = Yield strength d = average grain diameter o and k are constants for a metal. o Mpa and k = 0.74 Mpam 1/2 for mild steel.

36 Effects of Plastic Deformation Plastic deformation results in shearing of grains relative to each other. The grains elongate in rolling direction. Dislocations get rearranged. Grain structure at different regions of cartridge brass rolled into a wedge 6-40

37 Effect of Cold Work on Tensile Strength Number of dislocations are increased by cold work. Dislocation movements are hindered by both grain boundaries and other dislocations Strain Hardening Stress-Strain curves of 1018 steel 1018-Cold Rolled 1018-Annealed Figure

38 Solid Solution Strengthening Addition of one or more metals can increase the strength of metals. Solute atoms, in the case of substitutional solid solution, create stress fields around themselves and hinder dislocation movement. Distortion of lattice and clustering of like atoms also impedes dislocation movement. Example: Solid solution of 70 wt % Cu & 30 wt % Zn (cartridge brass) has tensile strength of 500 MPa. Tensile strength of unalloyed copper is 330 MPa 6-42

39 Superplasticity in Metals At elevated temperature and slow loading, some alloys deform 2000%. Annealed Ti alloy Elongates 12% at room temperature Elongates up to 1170% at 870 o C and 1.3x10 -4 /s loading rate. Conditions: very fine grain size (5-10 microns) * Highly strain sensitive * Temperature above 0.5 Tm * Slow strain rate

40 Mechanism of Superplasticity Very limited dislocation activity Deformation mechanism: Grain boundary sliding Grain boundary diffusion Sliding and rotation of individual grains. Applications: Metal forming operations. Blow forming to produce automobile hoods. Grains before and after deformation

41 Nanocrystalline Metals Average grain diameter < 100 nm Results in high strength and hardness, and Superplasticity If grain diameter reduces from 10 microns to 10 nm, yield strength of copper increases 31 times. Very difficult to produce nanocrystalline metals. If d < 5 nm, elastic modulus drops as more atoms are in grain boundary Hall-Petch equation is invalid in lower nanocrystalline range. Negative Hall-Petch effect might take place

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